Lecture Notes on Algebraic Topology
Jie Wu
Department of Mathematics, National University of Singapore, Singapore 119260,
Republic of Singapore,
URL: www.math.nus.edu.sg/~matwujie
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Contents
Chapter 1. Introduction
1. Sets
2. Monoids and Groups
3. G-sets
4. Categories and Functors
5
5
7
9
10
Chapter 2. General Topology
1. Metric spaces
2. Topological Spaces
3. Continuous Functions
4. Induced Topology
5. Quotient Topology
6. Product Spaces, Wedges and Smash Products
7. Topological Groups and Orbit Spaces
8. Compact Spaces, Hausdorff Spaces and Locally Compact Spaces
9. Mapping Spaces and Compact-open Topology
10. Manifolds and Configuration Spaces
13
13
14
16
17
19
20
23
25
31
37
Chapter 3. Elementary Homotopy Theory
1. Homotopy Sets
2. Homotopy Equivalences and Contractible Spaces
3. Retraction, Deformation and Homotopy Extension Property
4. H-spaces and Co-H-spaces
5. Barratt-Puppe Exact Sequences
41
41
46
47
52
65
Chapter 4. The Fundamental Groups and Covering Spaces
1. The fundamental Group
2. The Seifert-Van Kampen Theorem
3. Covering Spaces
4. The Lifting Theorem For Covering Spaces
5. Universal Covering
71
71
77
84
87
92
Chapter 5. Homology
1. Eilenberg-Steenrod Axioms
2. Computations and Applications
3. ∆-sets, Simplicial Sets and Homology
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99
106
114
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CONTENTS
Chapter 6. Some Suggested Topics for Your Further Reading/Study
1. Kă
unneth Theorem, Cohomology and Hopf Algebras
2. Hurewicz Theorem and Whitehead Theorem
3. Fibrations and Fibre Sequences
4. Spectral Sequences
5. Localization and Completion of Spaces
6. Simplicial Homotopy Theory and Simplicial Groups
7. Configuration Spaces and Combinatorial Models for Mapping Spaces
8. Homotopy Decompositions of Spaces
9. Cohen Groups
10. Homotopy Groups and the Exponent Problem in Homotopy Theory
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125
131
134
168
168
168
168
168
168
168
Bibliography
169
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CHAPTER 1
Introduction
1. Sets
Let X and Y be sets. The notation Y ⊆ X means that Y is a subset of X and Y ⊂ X means
that Y is a proper subset of X, that is Y ⊆ X and Y = X. Let X \ Y denote the set
X \ Y = {x|x ∈ X
and x ∈ Y }.
The empty set is denoted by ∅.
Let X and Y be sets. The Cartesian product is defined by
X × Y = {(x, y)|x ∈ X, y ∈ Y }.
Note: If X and Y are finite sets of m and n elements, respectively, then X × Y is a finite set of
mn elements.
Let X1 , · · · , Xn be sets. The Cartesian product is defined by
X1 × X2 × · · · × Xn = {(x1 , x2 , · · · , xn )|xi ∈ Xi , 1 ≤ i ≤ n}.
The infinite Cartesian product is defined similarly. For example, let {Xα |α ∈ I} be a family of sets.
Then
Xα = {(xα )|xα ∈ Xα , α ∈ I}.
α∈I
The α-coordinate projection
Xα → Xα
πα :
α ∈I
is defined by
πα ((xα )) = xα .
Theorem 1.1. Let {Xα |α ∈ I} be a family of set. Then the Cartesian product
the following universal lifting property:
α∈I
Xα satisfies
Let X be any set and let fα : X → Xα be any function for each α ∈ I. Then there is a
unique function
f: X →
Xα
α∈I
such that
fα = πα ◦ f
for each α.
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1. INTRODUCTION
Proof. Let f : X →
α∈I
Xα be a function defined by
f (x) = (fα (x))
for each x ∈ X. Then f is a function with the property that fα (x) = πα ◦ f (x) for any x and so
fα = πα ◦ f . This shows the existence of the universal lifting property. Let g : X → α∈I Xα be
any function with the property that fα = πα ◦ g for each α. Then the α-th coordinate of g(x) is
fα (x) for each x ∈ X. Thus g = f defined above. This shows the uniqueness of the universal lifting
property.
Let f : X → Y be a function. Then the image of f is defined by
Im(f ) = f (X) = {y ∈ Y |y = f (x)
for
some
x ∈ X}.
The identity function on X is denoted by idX , id or 1. Thus id(x) = x.
Exercise 1.1. Let f : X → Y be a function. Let {Xα |α ∈ I} be a family of subsets of X. Then
1) show that
f(
Xα ) =
f (Xα );
α∈I
α∈I
2) show that
Xα ) ⊆
f(
α∈I
f (Xα );
α∈I
3) show by example that
f(
Xα ) =
α∈I
f (Xα )
α∈I
in general.
Let f : X → Y be a function. Let A be a subset of X. Then the restriction f |A : A → Y is the
function defined by
f |A (a) = f (a)
for a ∈ A. Let B be a subset of Y . The pre-image f −1 (B) is defined by
f −1 (B) = {x ∈ X|f (x) ∈ B}.
Note that f −1 (B) could be an empty set.
Exercise 1.2. Let f : X → Y be a function. Let {Bβ |β ∈ J} be a family of subsets of Y . Then
1) show that
f −1 (
Bβ ) =
f −1 (Bβ );
β∈J
β∈J
2) show that
f −1 (
β∈J
f −1 (Bβ );
Bβ ) =
β∈J
3) show that
f −1 (Y \ Bβ ) = X \ f −1 (Bβ ).
A function f : X → Y is said to be bijective if it is one-to-one and onto. In this case the inverse
is denoted by f −1 : Y → X. Note that f −1 is also bijective. If there is a bijective function f from
X to Y , we call that X is isomorphic to Y as sets.
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2. MONOIDS AND GROUPS
7
Exercise 1.3. Let X be a set. Let Xα be a family of sets with indices α in a set I. Suppose
that Xα = X for each α. Show that α∈I Xα is isomorphic to the set of functions from I to X.
A relation on a set X is a subset ∼ of X × X. We write x ∼ y if (x, y) ∈∼. A relation on X is
an equivalence relation if it satisfies
1) the reflexive condition: x ∼ x for all x ∈ X;
2) the symmetric condition: If x ∼ y, then y ∼ x;
3) the transitive condition: If x ∼ y and y ∼ z, then x ∼ z.
The equivalence class of x is the set
{x} = {y ∈ X|x ∼ y}.
Exercise 1.4. Let ∼ be an equivalence relation on X. Show that each element of X belongs to
exactly one equivalence class.
2. Monoids and Groups
A binary operation (multiplication) on a set X is a function à : X ì X X. We abbreviate
à(x, y) to xy or x + y. A monoid M is a set M together with a multiplication µ : M × M → M
satisfying the following conditions:
1) (identity) there exists an element 1 ∈ M such that
1x = x1 = x
for any x ∈ M ;
2) (associativity) the equation
(x1 x2 )x3 = x1 (x2 x3 )
holds for any x1 , x2 , x3 ∈ M .
A group is a monoid G satisfying
3) (inverse) For each x ∈ G, there exists an element x−1 ∈ G such that
xx−1 = x−1 x = 1.
In other words, a group is a monoid in which every element is invertible. Note that if x is invertible,
then the inverse of x is unique. A group (or monoid) G is said to abelian or commutative if xy = yx
for any x, y ∈ G. Let G and H be monoids (or groups). Then the Cartesian product G × H is a
monoid (or group) under the multiplication defined by
(g, h)(g , h ) = (gg , hh ).
In additive case, we write G ⊕ H for G × H.
A subset H of a group (monoid) is a subgroup (submonoid) of G if H is a group (monoid) under
the binary operation of G. Let H be a subgroup (submonoid) of G and let g ∈ G. The left and
right cosets of H by g are defined by
gH = {gh|h ∈ H}
Hg = {hg|h ∈ H}.
+
Example 1.2. Let Z be the set of non-negative integers. Then Z+ is a monoid under the
addition +. Z+ is a submonoid of Z. Z is often called the group completion of the monoid Z+ ,
i.e. the “smallest group” that contains Z+ . The set of natural numbers is a monoid under the
multiplication. The “group completion” of natural numbers is the set of positive rational numbers
with the multiplication.
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1. INTRODUCTION
In general, monoids and the “group completion” of monoids are very complicated and there are
many research papers about these topics.
Let G and H be monoids (or groups). A homomorphism f : G → H is a function such that
f (1) = 1 and
f (xy) = f (x)f (y)
for any x, y ∈ G.
Exercise 2.1. Let G and Hbe groups and let f : G → H be a function such that f (xy) =
f (x)f (y) for any x, y ∈ G. Show that
1) f (1) = 1;
2) f (x−1 ) = f (x)−1 for any x ∈ G.
Let G and H be monoids (or groups). The kernel of a homomorphism f : G → H is the set
Ker(f ) = {x ∈ G|f (x) = 1}.
Note that a homomorphism f is one-to-one (a monomorphism) if and only if
Ker(f ) = {1}.
A monoid (or group) G is called isomorphic to H if there is a bijective homomorphism f : G → H.
In this case, we write G ∼
= H.
= H or f : G ∼
A subgroup K of G is normal if gxg −1 ∈ K for all g ∈ G and x ∈ K. Let G and H be groups.
Then the kernel of a homomorphism f : G → H is a normal subgroup of G. The image of f is a
subgroup of H which is not normal in general.
Exercise 2.2. Let K be a normal subgroup of a group G. Show that
1) gK = Kg for any g ∈ G;
2) the set
G/K = {gK|g ∈ G}
is a group under the operation
(gK)(g K) = (gg )K.
The group G/K is called the quotient group of G by K.
Let G be a group and let g ∈ G. The subgroup generated by g is the subset
g = {g n |n ∈ Z}.
Proposition 1.3. Let G be a group and let g ∈ G. Then g is isomorphic to Z or Z/nZ for
some n.
Proof. Let φ : Z → g be the function defined by
φ(n) = g n .
Then φ is a homomorphism of groups because
φ(m + n) = g m+n = g m g n = φ(m)φ(n).
Note that φ is an epimorphism, that is φ is onto. If g m = 1 for any positive integer m, then φ is an
isomorphism. Suppose that g m = 1 for some positive integer m. Let
n = min{m|g m = 1, m > 0}.
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3. G-SETS
9
Then
Ker(φ) = nZ
and so
g ∼
= Z/nZ.
If G = g for some g, we say that G is a cyclic group with generator g. A set of generators for
a group G is a subset S of G such that each element in G is a product of powers of elements taken
from S. A group G is called finitely generated if it is generated by a finite subset.
A free abelian group of rank n is the direct sum
Z⊕n = Z ⊕ Z ⊕ · · · ⊕ Z.
Theorem 1.4 (Decomposition Theorem). Let G be a finitely generated abelian group. Then G
is isomorphic to
H 0 ⊕ H1 ⊕ H2 ⊕ · · · ⊕ Hm ,
where H0 is a free abelian group and Hi is a cyclic group of prime power order for 1 ≤ i ≤ m.
The proof can be found in any text book of algebra.
A commutator in a group G is an element
[g, h] = ghg −1 h−1 .
for some elements g, h ∈ G. The commutator subgroup [G, G] is the subgroup of G generated by
all commutators of G. The commutator subgroup [G, G] is normal. The group G/[G, G] is called
the abelianization of the group G. Note that a group G is abelian if and only if the commutator
subgroup [G, G] is trivial. A group G is called perfect if [G, G] = G. An example of perfect groups
is the alternating groups An for n > 4. Non-commutative groups are much more complicated than
abelian groups.
3. G-sets
Let G be a group. A set X is called a left G-set if there is an operation à : G ì X → X,
(g, x) → g · x, such that
1) 1 · x = x for all x ∈ X;
2) (gh) · x = g · (h · x) for all g, h ∈ G and x ∈ X.
A set X is called a right G-set if there is an operation à : X ì G X, (g, x) → x · g, such that
1) x · 1 = x for all x ∈ X;
2) x · (gh) = (x · g) · h for all g, h ∈ G and x ∈ X.
Example 1.5. Let H be a subgroup of a group G. Then the set of left cosets {gH|g ∈ G} is a
left G-set and the set of right cosets {Hg|g ∈ G} is a right G-set.
Theorem 1.6. Let X be a left G-set. For any g ∈ G, the function θg : X → X defined by
x→g·x
is a bijective.
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1. INTRODUCTION
Proof. From the definition, we have that θg θh = θgh and θ1 = idX . Thus
θg θg−1 = idX = θg−1 θg
and so θg is a bijective.
Similarly, if X is a right G-set, then the function θg : X → X defined by x → x · g is a
bijective.
4. Categories and Functors
A category may be thought of intuitively as consisting of sets, possibly with additional structure,
and functions, possibly preserving additional structure. More precisely, a category C consists of
1) A class of objects
2) For every ordered pair of objects X and Y , a set Hom(X, Y ) of morphisms with domain
f
X and range Y ; if f ∈ Hom(X, Y ), we write f : X → Y or X ✲ Y
3) For every ordered triple of objects X, Y and Z, a function associating to a pair of morphisms f : X → Y and g : Y → Z their composite
g◦f: X →Z
These satisfy the following two axioms:
Associativity. If f : X → Y , g : Y → Z and h : Z → W , then
h ◦ (g ◦ f ) = (h ◦ g) ◦ f : X → W.
Identity. For every object Y there is a morphism idY : Y → Y such that if f : X → Y ,
then idY ◦f = f , and if h : Y → Z, then h ◦ idY = h.
A category is said to be small if the class of objects is a set. The category of sets means the
category in which the objects are sets and the morphisms are functions. The category of sets is
NOT small. But there are many small categories. For instance, the category of finite sets, that is in
which the objects are finite sets and the morphisms are functions between finite sets. We list some
examples of categories:
1) The category of sets and functions.
2) The category of pointed sets (A pointed set means a non-empty set X with a base point
x0 ∈ X) and pointed functions (that is the functions that preserving the base points).
3) The category of finite ordered sets and monotone functions (that is f (x) ≤ f (y) is x ≤ y).
This category is usually denoted by ∆. The objects in ∆ are given by {0, 1, · · · , n} for
n ≥ 0 and the morphisms in ∆ are given by monotone function from {0, 1, · · · , m} to
{0, 1, · · · , n} for any m, n.
4) The category of groups and homomorphisms.
5) The category of monoids and homomorphisms.
6) The category of topological spaces and continuous functions. Topological space is a generalization of the usual spaces such as Euclidian spaces Rn , spheres, polyhedra, metric
spaces and etc. We will give the definition of topological space in the next chapter.
Let C be a category. A subcategory C ⊆ C is a category such that
a) The objects of C are also objects of C;
b) For objects X and Y of C , HomC (X , Y ) is a subset of HomC (X , Y ) and
c) If f : X → Y and g : Y → Z are morphisms of C , their composite in C equals their
composite in C.
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4. CATEGORIES AND FUNCTORS
11
C is called a full subcategory of C if C is a subcategory of C and for objects X and Y in
C , HomC (X , Y ) = HomC (X , Y ). For example, the category of groups and homomorphisms is a
subcategory of the category of sets and functions but it is not a full subcategory. The category of
finite sets and functions is a full subcategory of the category of sets and functions.
Let C be a category. A morphism f : X → Y is called an equivalence if there is a morphism
g : Y → X such that g ◦ f = idX and f ◦ g = idY .
Let C and D be categories. A covariant functor (or contravariant functor) T from C to D consists
of an object function which assigns to every object X of C an object T (X) of D and a morphism
function which assigns to every morphism f : X → Y of C a morphism T (f ) : T (X) → T (Y ) [or
T (f ) : T (Y ) → T (X)] of D such that
a) T (idX ) = idT (X) and
b) T (g ◦ f ) = T (g) ◦ T (f ) [or T (g ◦ f ) = T (f ) ◦ T (g)].
Theorem 1.7. Let T be a functor from a category C to a category D. Then T maps equivalences
in C to equivalences in D.
Proof. Assume that T is covariant (the argument is similar if T is contravariant). Let f : X →
Y be an equivalence and let f −1 : Y → X be its inverse. Since f −1 ◦ f = idX and f ◦ f −1 = idY ,
T (f −1 ) ◦ T (f ) = idT (X) and T (f ) ◦ T (f −1 ) = idT (Y ) . Thus T (f ) is an equivalence.
A topological problem on spaces is to how to classify topological spaces. In other words, roughly
speaking, how to know whether a space X is homeomorphic to another space Y or not. Basic ideas
in algebraic topology is to introduce various functors from the category of topological spaces to
“algebraic” categories such as the category of groups, the category of abelian groups, and the
category of modules and etc. Homology, fundamental group and higher homotopy groups are most
important functors from the category of spaces to the category of groups.
For example, we will know that the fundamental group of R2 \ {0} is Z but the fundamental
group of R \ {0} is {0}. By Theorem 1.7, we have that R \ {0} is not homeomorphic to R2 \ {0}
and so R is not homeomorphic to R2 . This is a simple example. Actually we will be able to classify
all of (2-dimensional) surfaces in this course using the fundamental group.
Let C be any category. A simplicial object over C means a contravariant functor X : ∆ → C.
A simplicial set means a simplicial object over the category of sets. Similarly, we have simplicial
groups, simplicial monoids, simplicial algebras and etc. Simplicial sets and simplicial groups are
combinatorial models for spaces and topological groups, respectively, in homotopy sense. By its
combinatorial means, we can say that homotopy theory studies “functors”. On the other hand, by
its geometric means, homotopy theory studies continuous deformations of spaces and continuous
maps.
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CHAPTER 2
General Topology
1. Metric spaces
Let X be a set. A metric d for X is a function d : X × X → R satisfying
1) d(x, y) = 0 if and only if x = y;
2) (triangle inequality)
d(x, y) + d(x, z) ≥ d(y, z).
In this case X is called a metric space with the metric d.
Proposition 2.1. If d is a metric for X, then d(x, y) ≥ 0 and d(x, y) = d(y, x) for all x, y ∈ X.
Proof. By the triangle inequality, we have
2d(x, y) = d(x, y) + d(x, y) ≥ d(y, y) = 0,
d(x, y) = d(x, y) + d(x, x) ≥ d(y, x),
d(y, x) = d(y, x) + d(y, y) ≥ d(x, y).
Thus d(x, y) ≥ 0 and d(x, y) = d(y, x).
a) Show that each of the following is a metric for Rn :
Exercise 1.1.
n
(xi − yi )2 )1/2 = ||x − y||;
d(x, y) = (
d(x, y) =
i=1
0
1
if
if
x = y,
x = y;
n
|xi − yi |;
d(x, y) =
d(x, y) = max1≤i≤n |xi − yi |.
i=1
b) Show that d(x, y) = x − y does not define a metric on R.
c) Show that d(x, y) = min1≤i≤n |xi − yi | does not define a metric on Rn .
d) Let d be a metric. Show that d defined by
d (x, y) =
d(x, y)
1 + d(x, y)
is also a metric.
Definition 2.2. Let (X, dX ) and (Y, dY ) be metric spaces. A function f : X → Y is said to be
continuous at x ∈ X if for any x > 0 there exists δx > 0 such that dY (f (x), f (y)) < x whenever
dX (x, y) < δx . The function f is said to be continuous if it is continuous at all points x ∈ X.
Exercise 1.2. Let X be a metric space with metric d. Let y ∈ X.Show that the function
f : X → R defined by f (x) = d(x, y) is continuous.
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2. GENERAL TOPOLOGY
Definition 2.3. A subset U of a metric space (X, d) is said to be open if for any x ∈ U there
exists x > 0 such that if y ∈ X and d(y, x) < x then y ∈ U .
In other words U is open if for any x ∈ U there exists an x > 0 such that the open ball
B x (x) = {y ∈ X|d(y, x) <
x}
⊆ U.
Exercise 1.3.
a) Show that B (x) is always an open set for any x and any
b) Which of the following subsets of R2 (with the usual metric) are open?
{(x, y)|x2 + y 2 < 1} ∪ {(1, 0)},
> 0.
{(x, y)|x2 + y 2 ≤ 1},
{(x, y)||x| < 1}, {(x, y)|x + y < 0},
{(x, y)|x + y ≥ 0}, {(x, y)|x + y = 0}.
Exercise 1.4. Show that if U is the family of open sets arising from a metric space then
i) The empty set ∅ and the whole set belong to U;
ii) The intersection of two members of U belongs to U;
iii) The union of any number of members of U belongs to U.
Theorem 2.4. A function f : X → Y between two metric spaces is continuous if and only if
for any open set U in Y the set f −1 (U ) is open in X.
Proof. Suppose that f is continuous and U is open in Y . Let x ∈ f −1 (U ). Then f (x) ∈ U .
Since U is open, there exists > 0 such that B (f (x)) ⊆ U . By the definition, there exists δ > 0
such that f (y) ∈ B (f (x)) whenever y ∈ Bδ (x), that is f (Bδ (x)) ⊆ B (f (x). Thus
Bδ (x) ⊆ f −1 (B (x)) ⊆ f −1 (U )
and so f −1 (U ) is open.
Conversely let x ∈ X; then B (f (x)) is an open subset of Y and so f −1 (B (f (x))) is an open
subset of X. Thus there exists δ > 0 with
Bδ (x) ⊆ f −1 (B (f (x))).
In other words dY (f (x), f (y)) <
whenever dX (x, y) < δ, that is f is continuous.
2. Topological Spaces
Definition 2.5. Let X be a set. A topology U for X is a collection of subsets of X satisfying
i) ∅ and X are in U;
ii) the intersection of two members of U is in U;
iii) the union of any number of members of U is in U.
The set X with U is called a topological space. The members U ∈ U are called the open sets.
Exercise 2.1. Let U be a topology for X. Show that the intersection of a finite number of
members of U is in U.
Note: The intersection of infinitely many open sets is called a Borel set which is not open in general.
Let X be a metric space and let U be the family of open sets. Then U is a topology. This
topology is called the metric topology. Note that two different metrics may give rise to the same
topology.
Exercise 2.2. Let X be a metric space with metric d. Let d be the new metric defined in
Exercise 1.1. Then (X, d) and (X, d ) has the same topology.
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2. TOPOLOGICAL SPACES
15
Given a set X there may be different choices of topologies for X.
Exercise 2.3. Let X = {a, b}. Show that there are four different topologies given as follows:
U1 = {∅, X}, U2 = {∅, {a}, X}, U3 = {∅, {b}, X}, U4 = {∅, {a}, {b}, X}.
Exercise 2.4. Let X be a set. Let U1 = {∅, X}, let U2 = S(X) be the set of all subsets of X
and let
U3 = {U ⊆ X|X \ U is finite}.
Show that U1 , U2 and U3 are topologies for X.
U1 is called indiscrete topology, U2 is called discrete topology and U3 is called finite complement
topology.
Let X be a topological space and let A be a subset of X. The largest open set contained in A,
◦
this is denoted by A and is called the interior of A. For example, let X = Rn . Then the interior of
the closed ball
Dr (x) = {y|d(x, y) ≤ r}
is the open ball Br (x) = {y|d(x, y) < r}.
Exercise 2.5. Let X = Rn with the usual topology. Let
n
n
I = [0, 1] × · · · × [0, 1] .
Show that
n
◦
n
I = (0, 1) × · · · × (0, 1) .
Let X be a topological space. A subset N ⊆ X with x ∈ N is called a neighborhood of x if there
is an open set U with x ∈ U ⊆ N . For example, if X is a metric space, then the closed ball D (x)
and the open ball B (x) are neighborhoods of x.
Exercise 2.6. Let X be a topological space. Prove each of the following statements.
a) For each point x ∈ X there is at least one neighborhood of x.
b) If N is a neighborhood of x and N ⊆ M then M is also a neighborhood of x.
c) If M and N are neighborhoods of x then so is N ∩ M .
d) For each x ∈ X and each neighborhood N of x there exists a neighborhood U of x such
that U ⊆ N and U is a neighborhood of each of its points.
Definition 2.6. A subset C of a topological space X is said to closed if X \ C is open.
Theorem 2.7.
i) ∅ and X are closed;
ii) the union of any pair of closed sets is closed;
iii) the intersection of any number of closed sets is closed.
Note: The union of infinitely many closed sets is not closed in general.
Exercise 2.7. Let X be a set and let V be a family of subsets of X satisfying
i) ∅, X ∈ V;
ii) the union of any pair of members of V belongs to V;
iii) the intersection of any number of members of V belongs to V.
Show that U = {X − V |V ∈ V} is a topology for X.
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16
2. GENERAL TOPOLOGY
Let Y be a subset of a topological space X. The set
Y¯ =
{F |F ⊇ Y
F
is
closed}
is called the closure of Y . The set Y = Y¯ \ Y is called the set of limit points of Y .
Proposition 2.8. Let Y be a subset of a topological space X. Then x ∈ Y¯ if and only if for
every neighborhood N of x, N ∩ Y = ∅.
Proof. Let x ∈ Y¯ and suppose that N is a neighborhood of xwith N ∩ Y = ∅. Then there is
an open neighborhood U of x with U ⊆ N . Thus X \ U is a closed set and Y ⊆ X \ U . It follows
that Y¯ ⊆ X \ U and so x ∈ U . One gets a contradiction.
Conversely suppose that x ∈ Y¯ . Since Y¯ is closed, X \ Y¯ is an (open) neighborhood of x so
that (X \ Y¯ ) ∩ Y = ∅ is a contradiction.
Exercise 2.8. Let X = R with the usual topology. Find the closure of each of the following
subsets of X:
A = {1, 2, 3, · · · }, B = {x|x
is
rational}, C = {x|x
is
irrational}.
Exercise 2.9. Prove each of the following statements.
a)
b)
c)
d)
If Y is a subset of a topological space X with Y ⊆ F ⊆ X and F is closed then Y¯ ⊆ F .
Y is closed if and only if Y = Y¯ .
Y¯ = Y¯ .
¯
A ∪ B = A¯ ∪ B.
e)
f)
g)
h)
i)
X\ Y = X \ Y .
Y¯ = Y ∪ ∂Y where ∂Y = Y¯ ∩ (X \ Y ) (∂Y is called the boundary of Y ).
Y is closed if and only if ∂Y ⊆ Y .
∂Y = ∅ if and only if Y is both open and closed.
For a < b ∈ R
∂(a, b) = ∂[a, b] = {a, b}.
◦
3. Continuous Functions
Definition 2.9. A function f : X → Y between two topological spaces is said to be continuous
if for every open set U of Y the preimage f −1 (U ) is open in X.
A continuous function from a topological space to a topological space is often simply called a
map. The category of topological spaces is defined as follows: the objects are topological spaces and
the morphisms are maps, that is continuous functions.
Theorem 2.10. Let X and Y be topological spaces. A function f : X → Y is continuous if and
only if f −1 (C) is closed for any closed subset C of Y .
Proof. Suppose that f is continuous and let C be a closed set in Y . Then Y \ C is an open
set and so f −1 (Y \ C) = X \ f −1 (C) is an open set. It follows that f −1 (C) is a closed set. Now
suppose that f −1 (C) is closed for any closed set C and let U be an open set. Then Y \ U is a closed
set and so X \ f −1 (U ) = f −1 (Y \ U ). Thus f −1 (U ) is an open set.
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4. INDUCED TOPOLOGY
17
So far we have two general methods to see whether a function is continuous or not, that is by
the definition or by the theorem above. If f : X → Y is a function between metric spaces, then we
can also use − δ method to test whether f is continuous or not. As we know in calculus that the
compositions of continuous functions is still continuous. This is actually true in general.
Theorem 2.11. Let X, Y and Z be topological spaces. If f : X → Y and g : Y → Z are
continuous functions then the composite g ◦ f : X → Z is continuous.
f
−1
Proof. Let U be any open set in Z. Then g −1 (U ) is an open set in Y and so (g ◦ f )−1 (U ) =
(g −1 (U )) is an open set in X.
Definition 2.12. Let X and Y be topological spaces. We say that X and Y are homeomorphic
if there exist continuous functions f : X → Y, g : Y → X such that f ◦ g = idY and g ◦ f = idX . We
write X ∼
= Y and say that f and g are homeomorphisms between X and Y .
By the definition, a function f : X → Y is a homeomorphism if and only if
i) f is a bijective;
ii) f is continuous and
iii) f −1 is also continuous.
Equivalently f is a homeomorphism if and only if 1) f is a bijective, 2) f is continuous and 3) f is
an open map, that is f sends open sets to open sets. Thus a homeomorphism between X and Y is
a bijective between the points and the open sets of X and Y .
A very general question in topology is how to classify topological spaces under homeomorphisms.
For example, we know (from complex analysis and others) that any simple closed loop is homeomorphic to the unit circle S 1 . Roughly speaking topological classification of curves is known. The
topological classification of (two-dimensional) surfaces is known as well. However the topological
classification of 3-dimensional manifolds (we will learn manifolds later.) is quite open. The famous
Poicar´e conjecture is related to this problem.
Exercise 3.1. Give an example of spaces X, Y and a continuous bijective f : X → Y such that
f −1 is NOT continuous. (Hint: Give a set X. Look at the discrete topology, the indiscrete topology
and the identity function.)
A pointed space means a topological space X together with a point x0 ∈ X. The point x0 is
called the base point of X. We often write ∗ for x0 . Let X and Y be pointed spaces with base points
x0 and y0 , respectively. A map f : X → Y is called a pointed map if f (x0 ) = y0 . The category
of pointed topological spaces means a category in which the objects are pointed spaces and the
morphisms are pointed maps.
4. Induced Topology
Definition 2.13. Let X be a topological space and let S be a subset of X. The topology on
S induced by the topology of X is the family of the sets of the form U ∩ S where U is an open set
in X. We call that the subset S with induced topology is a subspace of X.
Note: By this definition, an open set V in S means V = U ∩ S for some open set U in X. The
induced topology is also called the subspace topology.
Exercise 4.1. Let X be a topological space with the topology U and let S be a subset of X.
Show that
U ∩ S = {U ∩ S|U ∈ U }
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18
2. GENERAL TOPOLOGY
is a topology for S.
Example 2.14. Let S n be the n-sphere, that is,
n+1
S n = {x = (x1 , · · · , xn+1 ) ∈ Rn+1 |
x2i = 1} ⊆ Rn+1
i=1
with the induced topology. Then S n is a (closed) subspace of Rn+1 . Note that S n = Rn+1 ∩ S n is
an open set in S n but S n is NOT open in Rn+1 .
Proposition 2.15. Let S be a subspace of a topological space X. Then the inclusion function
i : S → X is continuous.
Proof. Let U be an open set in X. Then i−1 (U ) = U ∩ S is an open set in S.
Note: One can show that the subspace topology is the smallest topology such that the inclusion is
continuous.
Proposition 2.16. Let S be a subspace of a topological space X. Then
1) If S is open in X, then any open set in the subspace S is open in X;
2) If S is closed in X, then any closed set in the subspace S is closed in X.
Proof. The proofs of 1) and 2) are more or less identical. We only prove assertion 2). Let V
be a closed set in S. Then S \ V is an open set in S. By the definition, there is an open set U in
X such that
S \ V = U ∩ S.
Thus V = (X \ U ) ∩ S. Since S and X \ U are closed, V is closed.
Exercise 4.2. Show that
1) the subspace (a, b) of R is homeomorphic to R. (Hint: Use functions like x → tan(π(cx +
d)) for suitable c and d.)
2) the subspaces (1, ∞), (0, 1) of R are homeomorphic. (Hint: x → 1/x.)
3) S n \ {(0, 0, · · · , 0, 1)} is homeomorphic to Rn with the usual topology. (Hint: Define
φ : S n \ {(0, 0, · · · , 0, 1)} → Rn by
x2
xn
x1
,
··· ,
)
φ(x1 , x2 , · · · , xn+1 ) = (
1 − xn+1 1 − xn+1
1 − xn+1
and ψ : Rn → S n \ {(0, 0, · · · , 0, 1)} by
1
(2x1 , 2x2 , · · · , 2xn , x
ψ(x1 , · · · , xn ) =
1+ x 2
2
− 1).)
A map f : X → Y is called an embedding if f is one-to-one and X is homeomorphic to the image
f (X) with the subspace topology. The embedding problem in topology is as follows:
Given a topological space X. Can we embed X into Rn for some n? If not, can we
embed X into a Hilbert space? If yes, what is the minimal number n such that X can be
embedded in Rn ? This number is called the embedding number of X.
This question is important (and difficult in general) because a topological space X could be very
abstract but the spaces Rn are much easier to be understood. For instance, the circle S 1 can embed
in R2 but S 1 can not embed in R1 . Thus the embedding number of S 1 is 2. Well sometimes a space
X could be very simple but it could have a very complicated embedding in Rn .
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5. QUOTIENT TOPOLOGY
19
A knot K is a subspace of R3 that is homeomorphic to the circle S 1 . Two knots K1 and K2
are similar if there is a homeomorphism h : R3 → R3 such that h(K1 ) = K2 . The knot theory is to
study the classification of knots under this relation.
5. Quotient Topology
Definition 2.17. Let f : X → Y be a surjective function from a topological space X to a set
Y . The quotient topology on Y with respect to f is the family
Uf = {U |f −1 (U )
is
open in X}.
Exercise 5.1. Show that Uf above is a topology for Y .
Note: After giving the quotient topology on Y the function f : X → Y is continuous.
Example 2.18 (Projective Spaces). Let set RP n and CP n are defined as follows:
RP n = {l|l
is
a
line
in Rn+1
n
with 0 ∈ l},
n+1
CP = {l|l is a complex line in C
with 0 ∈ l}.
The topologies in RP n and CP n are given by the quotient topology under the quotient maps
Rn+1 \ {0} → RP n and Cn+1 \ {0} → CP n , respectively.
By this example, one can see that the quotient space Y could be much more complicated than
the original space X. The following theorem gives a general method to see whether a function from
Y to another space is continuous or not.
Theorem 2.19. Let X be a topological space and let f : X → Y be a surjective. Suppose that Y
are given the quotient topology with respect to f . Then a function g : Y → Z from Y to a topological
space Z is continuous if and only if the composite g ◦ f is continuous.
Proof. Suppose that g is continuous. Since f is continuous, the composite g ◦ f is continuous.
Now suppose that the composite g ◦ f is continuous. Let U be any open set in Z. Then
f −1 (g −1 (U )) = (g ◦ f )−1 (U )
is open in X and so g −1 (U ) is open in Y by the definition of quotient topology.
Exercise 5.2. Show that
1) RP 1 ∼
= S1;
1 ∼ 2
2) CP = S .
The famous Hopf fibration is the composite
✲ CP 1 ∼
S 3 ⊂ ✲ C2 \ {0} ✲
= S2.
Let A be a subspace of a space X. The space X/A is the quotient space
X/A = X/ ∼,
where ∼ is the equivalence relation generated by
a∼b
for any a, b ∈ A. As a set X/A = (X \ A) ∪ {∗}, where ∗ is the equivalence class of any particular
choice of elements in A. The topology in X/A is given by the quotient topology. Roughly speaking
X/A is the quotient space X by pinching out A to be one point.
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20
2. GENERAL TOPOLOGY
Exercise 5.3. Show that Dn /S n−1 is homeomorphic to S n .
The canonical inclusions Rn → Rn+1 , Cn → Cn+1 given by (x1 , · · · , xn ) → (x1 , · · · , xn , 0)
induce the maps RP n → RP n+1 and CP n → CP n+1 , respectively. Thus RP n and CP n+1 can be
considered as the subspaces of RP n+1 and CP n+1 , respectively, for each n.
Exercise 5.4. Show that RP n+1 /RP n ∼
= S n+1 and CP n+1 /CP n ∼
= S 2n+1 .
A fibrewise topological space means a map f : X → Y . In the case where f is an onto, it often
called a bundle. For each y ∈ Y , the subspace f −1 (y) ⊆ X is called the fibre at y. Let f : X → Y
be a bundle. Then
f −1 (y)
X=
y∈Y
and so X can be considered as the union of subspaces f −1 (y) with indexes in a topological space Y .
Fibre bundles and covering spaces are special bundles. We will study covering spaces in the next
chapter. The category of fibrewise topological spaces is a category in which the objects are fibrewise
topological spaces and the morphisms are given by the commutative diagrams
X
φ ✲
X
f
f
❄
Y
ψ ✲ ❄
Y .
In other words, the working objects for fibrewise topology are continuous maps and the “relations”
between the working objects are the diagrams above. One finds surprisingly that many results in the
homotopy theory of topological spaces also holds for the homotopy theory of fibrewise topological
spaces. Well the latter one is much more “abstract”.
6. Product Spaces, Wedges and Smash Products
Let X and Y be topological spaces with topologies UX and UY , respectively. Let
UX×Y = {
Uα × Vα ⊆ X × Y |Uα ∈ UX , Vα ∈ UY },
α
that is any member in UX×Y is the union of Cartesian products of open sets of X and Y .
Exercise 6.1. Let X and Y be topological spaces. Show that UX×Y is a topology for X × Y .
Definition 2.20. Let X and Y be topological spaces. The (Cartesian) product X × Y is the
set X × Y with the topology UX×Y .
Exercise 6.2. Show that R2 with the usual topology is the Cartesian product R1 × R1 .
Theorem 2.21. Let X ×Y be the Cartesian product of spaces X and Y . Then a set W ⊆ X ×Y
is open if and only if for any w ∈ W there exist Uw and Vw such that Uw is open in X, Vw is open
in Y and w ∈ Uw × Vw ⊆ W .
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6. PRODUCT SPACES, WEDGES AND SMASH PRODUCTS
21
Proof. Let W be an open set in X × Y and let w ∈ W . Then W = α Uα × Vα , where Uα
and Vα are open in X and Y , respectively. Thus there exists an index α such that w ∈ Uα × Vα .
Choose Uw = Uα and Vw = Vα . Conversely let w run over all elements in W we have
Uw × Vw
W =
w∈W
and so W is open.
Let πX : X ×Y → X, (x, y) → x, and πY : X ×Y → Y, (x, y) → y, be the coordinate projections.
−1
Since πX
(U ) = U × Y and πY−1 (V ) = X × V , the coordinate projections πX and πY are continuous.
Let f : Z → X and g : Z → Y be any maps from a space Z to X and Y , respectively. Let
φ : Z → X × Y be the function defined by φ(z) = (f (z), g(z). Then φ is the unique function such
that πX ◦ φ = f and πY ◦ φ = g.
Lemma 2.22. The function φ defined above is continuous.
Proof. Let U and V be open sets in X and Y , respectively. Then φ−1 (U × V ) = {z|f (z) ∈
U, g(z) ∈ V } = f −1 (U ) ∩ g −1 (V ) is an open set in Z. Now consider any open set W in X × Y .
Let z be any element in φ−1 (W ) and let w = φ(z). There exist open sets Uw and Vw such that
w ∈ Uw × Vw ⊆ W . Thus z ∈ φ−1 (Uw × Vw ) ⊆ φ−1 (W ) and φ−1 (W ) is a neighborhood of each of
its points. It follows that φ−1 (W ) is open.
By using the categorical language, Lemma 2.22 shows
Theorem 2.23. Let X and Y be topological spaces. Then Cartesian product X × Y is the
product of X and Y in the category of topological spaces.
Theorem 2.24. For any y ∈ Y , the subspace X × {y} ⊆ X × Y is homeomorphic to X.
Proof. Let f : X × {y} → X be the function defined by f (x, y) = x. Since f is the composite
f : X × {y}
⊂
✲ X ×Y
πX
✲ X,
the function f is continuous. Clearly f is a bijective. It suffices to show that f is an open map,
that is f sends open sets to open sets. Suppose that W is an open set in X × {y}. Then
Uα × Vα ) ∩ X × {y}
W =(
α
for some open sets Uα and Vα in X and Y , respectively. It follows that
f (W ) =
Uα
α,y∈Vα
is open.
Now we look at “infinite” Cartesian products. Let {Xα |α ∈ J} be a set of topological spaces.
Recall that the Cartesian product α∈J Xα of the sets Xα is the set of collections of elements (xα ),
one element xα in each Xα . Now An open set in α∈J Xα is defined to be the any union of the
following sets
Uα1 ,··· ,αn = {(xα )|xα1 ∈ Uα1 , · · · , xαn ∈ Uαn },
where a1 , · · · , an is any finite set of elements of J. This gives the topology on the product
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α∈J
Xα .
22
2. GENERAL TOPOLOGY
Proposition 2.25. The product topology on α∈J Xα is the smallest topology such that each
coordinate projection
πα :
Xα → Xα
α ∈J
is continuous.
Proof. Let V be a topology on α ∈J Xα such that each coordinate projection πα is continuous. Let Uα be an open set in Xα . Then
π −1 (Uα ) = {(xα )|xα ∈ Uα } ∈ V.
(1)
Since the product topology is given by the any union of any finite intersections of the sets of the
forms 1, it follows that the product topology is smaller than V.
Let X and Y be pointed spaces with base points x0 and y0 , respectively. Then the wedge X ∨ Y
of X and Y is defined to be the quotient space
(X
Y )/{x0 , y0 }.
The topology in X ∨Y is given by the quotient topology under the quotient map q : X Y → X ∨Y .
This topology can be described as follows. A subset U in X ∨ Y is open if and only if q −1 (U ) is
open. There are two cases. If ∗ ∈ U , then q −1 (U ) is either an open set in X that does not contain
x0 or an open set in Y that does not contain y0 . If ∗ ∈ U , then q −1 (U ) = U1 U2 for some open
set U1 in X that contains x0 and some open set U2 in Y that contains y0 . Thus
UX∨Y = {q(U )|x0 ∈ U ∈ UX } ∪ {q(V )|y0 ∈ V ∈ UY } ∪ {q(U1
❛
U2 )|x0 ∈ U1 ∈ UX , y0 ∈ U2 ∈ UY }.
Proposition 2.26. Let X and Y be pointed spaces with base points x0 and y0 , respectively.
Then X ∨ Y is homeomorphic to the subspace (X × {y0 }) ∪ ({x0 } × Y }) ⊆ X × Y.
Proof. Let Z = (X × {y0 }) ∪ ({x0 } × Y ) be the subspace of X × Y . Let fX : X → X × Y and
fY : Y → X × Y be the maps defined by fX (x) = (x, y0 ) and fY (y) = (x0 , y). The there is a unique
map φ : X ∨ Y → Z such that φ ◦ fX = iX and φ ◦ fY = iY . Clearly φ is bijective. It suffices to
show that φ is an open map. If U ∈ UX with x0 ∈ U , then φ(U ) = Z ∩ (U × Y ) is open in Z. If
V ∈ UY with y0 ∈ V , then φ(V ) = Z ∩ (X × V ) is open in Z. If U = q(U1 U2 ) with x0 ∈ U1 ∈ UX
and y0 ∈ U2 ∈ UY , then φ(U ) = Z ∩ (U1 × U2 ). Thus q is an open map.
Let X and Y be pointed spaces. The smash product X ∧ Y is defined by
(X × Y )/((X × {y0 }) ∪ ({x0 } × Y )).
We write x ∧ y for elements in X ∧ Y , where x ∈ X and y ∈ Y .
Theorem 2.27. Given three pointed spaces X, Y and Z, (X ∨ Y ) ∧ Z is homeomorphic to
(X ∧ Z) ∨ (Y ∧ Z).
Proof. The function f : X × Y × Z → X × Z × Y × Z, defined by f (x, y, z) = (x, z, y, z), is
clearly continuous. Let g be the composite
g: X × Y × Z
f
✲ X ×Z ×Y ×Z
proj.
✲ (X ∧ Z) × (Y ∧ Z).
Then
g((X ∨ Y ) × Z) ⊆ (X ∧ Z) ∨ (Y ∧ Z).
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7. TOPOLOGICAL GROUPS AND ORBIT SPACES
23
Moreover, the map g sends (X ∨ Y ) ∨ Z to the base point, so that g induces a map
g˜ : (X ∨ Y ) ∧ Z → (X ∧ Z) ∨ (Y ∧ Z),
where g˜((x, y0 ) ∧ z) = x ∧ z in X ∧ Z and g((x0 , y) ∧ z) = y ∧ z in Y ∧ Z.
Conversely, let h : (X ∧ Z) ∨ (Y ∧ Z) → (X ∨ Y ) ∧ Z be the map such that h|X∧Z and h|Y ∧Z
are the inclusions X ∧ Z ⊂ ✲ (X ∧ Y ) ∧ Z and Y ∧ Z ⊂ ✲ (X ∧ Y ) ∧ Z, respectively. Then
h(x ∧ z) = (x, y0 ) ∧ z and h(y ∧ z) = (x0 , y) ∧ z) so that g˜ ◦ h and h ◦ g˜ are identities, and hence g˜
is a homeomorphism.
Exercise 6.3. Show that S n ∧ S m ∼
= S n+m for any n, m.
7. Topological Groups and Orbit Spaces
A pointed topological space X is called an H-space of there is a continuous multiplication
µ : X × X → X, (x, y) → xy, such that x0 x = xx0 = x. The base point x0 is often denoted as ∗ or
1. Equivalently, a pointed space X is an H-space if and only if there is a map à : X ì X X such
that µ|X∨X = ∇, where ∇ : X ∨ X → X is the fold map defined by ∇(x, x0 ) = x and ∇(x0 , x) = x.
An H-space is called associative if diagram
X ìX ìX
à ì id
X
X ìX
idX ×µ
❄
X ×X
µ
µ
❄
✲ X
is commutative. An associative H-space is called a topological monoid. In other words, a topological
monoid is monoid as a set such that the multiplication is continuous. A topological group G means
a topological monoid such that there is a map ν : G → G, x → x−1 , with xx−1 = 1 = x−1 x, that is
the inverse is a continuous function.
Let X be a space and let G be a topological group. We say that G acts on X and that X is a
G-space if there is map à : G ì X X, denoted by (g, x) → g · x, such that
i) 1 · x = x for all x ∈ X;
ii) g · (h · x) = (gh) · x for all x ∈ X and g, h ∈ G, that is the diagram
GìGìX
idG ìà
GìX
àG ì idX
GìX
à
à
X
commutes.
Theorem 2.28. Suppose that X is a G-space. Then the function θg : X → X given by x → g·x is
a homeomorphism. It follows that there is a homomorphism from G to the group of homeomorphisms
of X.
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24
2. GENERAL TOPOLOGY
Proof. The function θg is the composite
X∼
= {g} × X ⊆ G × X
µ
✲ X.
Thus θg is continuous. From the definition of G-space we see that θg ◦ θh = θgh and θ1 = idX .
Thus θg ◦ θg−1 = idX = θg−1 ◦ θg and so θg is a homeomorphism. Now the function g → θg is a
homomorphism from G to the group of homeomorphisms of X.
Let X be a G-space. We can define an equivalence relation ∼ on X by
x ∼ y ⇔ g · x = y for some g ∈ G.
The quotient space X/ ∼, denoted by X/G, with the quotient topology is called the quotient space
of X by G.
Example 2.29.
1) Let G = Z/2 = {±1} with discrete topology and let X = S n . The
G-action on X is given by ±1 · x = ±x. Then S n /(Z/2) ∼
= RP n .
2) Let G = Z with the discrete topology and let X = R. The action of G on R is given by
n · x = n + x. Then R/Z ∼
= S1.
1
3) Let G = S ⊆ C. Then G is a topological group under the multiplication. Let S 2n−1 ⊆
R2n = Cn be the unit sphere. Let G act on S 2n−1 by
α · (z1 , z2 , · · · , zn ) = (αz1 , αz2 , · · · , αzn ).
Then S 2n−1 /S 1 ∼
= CP n .
2
4) Let Mn be the set of n × n-matrices over R. Then Mn = Rn is a topological space. Let
GL(n, R) = {A ∈ Mn | det(A) = 0} ⊆ Mn
with the subspace topology. Then GL(n, R) is a topological group, which called the general
linear group.
5) Let O(n) be the group of (real) orthogonal n × n matrices. O(n) is regarded as a subspace
2
of Rn with the subspace topology. For k ≤ n O(k) is regarded as the set of matrices of
the form
A
0
0 In−k
with A an orthogonal k × k-matrix and In−k the (n − k) × (n − k) identity matrix. Then
O(k) is a topological subgroup of O(n). In O(n) we also have the subgroup SO(n) of
orthogonal matrices with determinant 1, that is SO(n) is the kernel of det : O(n) → Z/2.
2
6) Let U (n) denote the group of n × n unitary matrices regarded as a subspace of Cn . We
have the inclusions
U (1) ⊆ U (2) ⊆ U (3) ⊆ · · · ⊆ U (n) ⊆ · · · .
Thus U (k) is a topological subgroup of U (n) for k ≤ n. We also have the subgroup
SU (n) ⊆ U (n) of n × n unitary matrices with determinant 1, that is SU (n) is the kernel
of det : U (n) → S 1 .
Theorem 2.30. Suppose that X is a G-space. Then the canonical projection π : X → X/G is
an open mapping.
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8. COMPACT SPACES, HAUSDORFF SPACES AND LOCALLY COMPACT SPACES
25
Proof. Let U be an open set in X. Then
π −1 (π(U )) = {x ∈ X|π(x) ∈ π(U )}
= {x ∈ X|x = g · y for some y ∈ U some g ∈ G} =
g · U.
g∈G
Since θg : X → X is a homeomorphism for each g ∈ G, g · U is open for each g then so π −1 (π(U ))
is open and hence π(U ) is open in X/G.
Exercise 7.1.
subspace
1) Let X be a G-space and define the stabilizer of x ∈ X to be the
Gx = {g ∈ G|g · x = x}
of G. Show that Gx is a topological subgroup of G.
2) Let X be a G-space and define the orbit of x ∈ X to be the subspace
G · x = {g · x|g ∈ G}
of X. Prove that G · x and G · y are either disjoint or equal for any x, y ∈ X.
8. Compact Spaces, Hausdorff Spaces and Locally Compact Spaces
Let X be a space. A cover of a subset S is a collection of subsets {Uj |j ∈ J} of X such that
S⊆
Uj .
j∈J
A cover is called finite if the indexing set J is finite. Let {Uj |j ∈ J} and {Vk |k ∈ K} be covers of
the subset S of X. {Uj |j ∈ J} is called a subcover of {Vk |k ∈ K} if
{Uj |j ∈ J} ⊆ {Vk |k ∈ K}.
Definition 2.31. Let X be a space. A subset S is called to be compact if every open cover of
S has a finite subcover. In particular, a space X is compact if every open cover of X has a finite
subcover.
Exercise 8.1. Show that a subset S of a space X is compact if and only if it is compact as a
space given the induced topology.
Exercise 8.2. Show that [0, 1] ⊆ R is compact.
The following theorem is useful.
Theorem 2.32. Let f : X → Y be a map. If S ⊆ X is a compact subspace, then f (S) is
compact.
Proof. Suppose that {Uj |j ∈ J} be an open cover of f (S). Then {f −1 (Uj )|j ∈ J} is an open
cover of S. Since S is compact, there exists a finite subset K of J such that
S⊆
{f −1 (Uk )|k ∈ K}.
But f (f −1 (Uk )) ⊆ Uk and so
f (S) ⊆
{f (f −1 (Uk ))|k ∈ K} ⊆ {Uk |k ∈ K}
which is a finite subcover of {Uj |j ∈ J}.
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